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3 years ago

CaCl2 + Na2CO3 → CaCO3 + 2 NaCl

Chemistry

2 answers:

alukav5142 [94]3 years ago

4 0

Answer:

8.45 moles

Explanation:

The balanced equation for the reaction is given below:

CaCl2 + Na2CO3 → CaCO3 + 2NaCl

From the balanced equation above,

1 mole of calcium chloride (CaCl2) produced 1 mole of calcium carbonate (CaCO3).

Therefore, 8.45 moles of calcium chloride (CaCl2) will also produce 8.45 moles of calcium carbonate (CaCO3)

From the illustration above, 8.45 moles of calcium carbonate (CaCO3) are produced.

Nana76 [90]3 years ago

3 0

Answer:

8.45 moles are produced

Explanation:

CaCl₂ + Na₂CO₃ → CaCO₃ + 2 NaCl

From the equation, we can see that for every 1 mole of CaCl₂ and 1 mole Na₂CO₃ will give 1 mole of CaCO₃ and 2 moles of NaCl

to calculate how many moles of CaCO₃ ,we simply multiply multiply each by the 8.45 moles of CaCl₂ which will reacts

these is because for every 1 mole of CaCl₂ and 1 mole Na₂CO₃ will give 1 mole of CaCO₃ and 2 moles of NaCl

therefore we have every 1x8.45(8.45) mole of CaCl₂ and 1x8.45(8.45) mole Na₂CO₃ will give 1x8.45(8.45) mole of CaCO₃ and 2x8.45(16.9) moles of NaCl

8.45 moles are produced in the reaction

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3 years ago

Consider the following reaction: 2 Bi(s) + 3 Cl2(g) → 2 BiCl3(s)

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Taking into account the reaction stoichiometry, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Bi + 3 Cl₂ → 2 BiCl₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Bi: 2 moles
Cl₂: 3 moles
BiCl₃: 2 moles

The molar mass of the compounds is:

Bi: 209 g/mole
Cl₂: 70.90 g/mole
BiCl₃: 315.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Bi: 2 moles ×209 g/mole=418 g

Cl₂: 3 moles ×70.90 g/mole= 212.7 g

BiCl₃: 2 moles ×315.45 g/mole= 630.9 g

<h3>Mass of Cl₂ required</h3>

The following rule of three can be applied: If by reaction stoichiometry 630.9 grams of BiCl₃ are formed by 212.7 grams of Cl₂, 3.32 grams of BiCl₃ are formed by how much mass of Cl₂?

$mass of Cl_{2} =\frac{3.32 grams of BiCl_{3}x212.7 grams of Cl_{2} }{630.9 grams of BiCl_{3}}$

<u><em>mass of Cl₂= 1.119 grams</em></u>

Finally, 1.119 grams of chlorine gas are required to produce 3.32 grams of bismuth chloride is produced.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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